Съдържание

Условие Отговори Решение

USAPhO 2005, Четвъртфинал — Задача 4

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2005

2005 Preliminary Exam — Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie / Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8\ N/kg$
Newton's gravitational constant	$G=6.67\times10^{-11}\ N\cdot m^2/kg^2$
Average Earth-Sun Distance	$1$ Astronomical Unit (A.U.) $=1.5\times10^{11}\ m$
Binomial expansion	$(1+x)^n=1+nx$ for $\vert x\vert \ll1$

Moment of Inertia about Center of Mass – Uniform Object (will not be provided on the second screening exam)

Object	Moment of Inertia
Disk	$\frac12MR^2$
Sphere	$\frac25MR^2$
Rod	$\frac1{12}ML^2$

Solutions to Problems

Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. Where alternate methods are presented, a student should be awarded points from only one method. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.) If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 2.b., if a student wrote down $mvb=mv_pR$ , they should get 10 points credit.

Условие

Top and side views of a disk pulled by a cord tangent to its rim

A thin disk of mass $M$ , radius $R$ , and height $H$ is initially at rest on a flat horizontal table as shown in top and side views above. There is no friction between the disk and the table. A long massless cord is wrapped around the disk and pulled with constant force $F$ parallel to the table.

a Find the ratio of rotational to translational kinetic energy. [15 т.]

b What is the total work done by the force $F$ during the disk's first revolution? [10 т.]

Отговори

Решение

a. The translational kinetic energy is $K_t=\frac12Mv^2$ or $K_t=W_F=Fd$ , [2] where $v$ is the time-dependent translational speed. From Newton's Second Law with $F$ the only force acting, the acceleration of the center of mass is given by $a=F/M$ . [1] Starting from rest, $v=at=Ft/M$ or $d=\frac12at^2=Ft^2/(2M)$ . [2] By either method, the translational kinetic energy is

K_t=\frac12M\left(\frac{Ft}{M}\right)^2=\frac{F^2t^2}{2M}\quad\text{or}\quad K_t=F\left(\frac{Ft^2}{2M}\right)=\frac{F^2t^2}{2M}.\qquad (4-1)

[1] The rotational kinetic energy is $K_r=\frac12I\omega^2$ or $K_r=\tau\theta$ , [2] where $\omega$ is the time-dependent rotational speed, $\theta$ is the angular displacement, $\tau$ is the torque, and $I$ the disk's rotational inertia about its center of mass: $I=\frac12MR^2$ . [1] The torque about the center of mass is $|\vec\tau|=|\vec R\times\vec F|=RF$ . [1] From Newton's Second Law for rotational motion the angular acceleration about the center of mass is given by

\alpha=\frac\tau I=\frac{RF}{\frac12MR^2}=\frac{2F}{MR}.\qquad (4-2)

[1] Starting from rest, $\omega=\alpha t=2Ft/(MR)$ or $\theta=\frac12\alpha t^2=Ft^2/(MR)$ . [2] By either method, the rotational kinetic energy is

K_r=\frac12\left(\frac12MR^2\right)\left(\frac{2Ft}{MR}\right)^2=\frac{F^2t^2}{M}\quad\text{or}\quad K_r=(RF)\frac{Ft^2}{MR}=\frac{F^2t^2}{M}.\qquad (4-3)

[1] Dividing (4-3) by (4-1),

\frac{K_r}{K_t}=\frac{F^2t^2/M}{F^2t^2/(2M)}=2.

[1]

Note:

\frac{K_r}{K_t}=\frac{\frac12I\omega^2}{\frac12Mv^2}=\frac{(\frac12MR^2)\omega^2}{Mv^2}=\frac12\left(\frac{\omega R}{v}\right)^2.

This is not rolling without slipping and $v\ne\omega R$ . The answer $\frac12$ is only worth 4 points.

b. The net work done by the force is equal to the change in kinetic energy: $W=\Delta K_t+\Delta K_r$ . [2] Since the disk starts at rest both initial kinetic energies are zero and

W=K_t+K_r=\frac{F^2t^2}{2M}+\frac{F^2t^2}{M}=\frac{3F^2t^2}{2M},\qquad (4-4)

[3] where $t$ is the time it takes to make the first rotation. With uniform angular acceleration and starting from rest,

\theta=\frac12\alpha t^2.\qquad (4-5)

[1] From (4-2), $\alpha=2F/(MR)$ , [1] and for a single rotation $\theta=2\pi$ . [1] Solving (4-5) for $t^2$ ,

t^2=\frac{2\theta}{\alpha}=\frac{2(2\pi)}{2F/(MR)}=\frac{2\pi MR}{F}.

[1] Substituting into (4-4),

W=\frac{3F^2t^2}{2M}=\frac{3F^2}{2M}\left(\frac{2\pi MR}{F}\right)=3\pi FR.

[1]

Теми и трудност

Трудност: неоценена.

Механика · Въртене на твърдо тяло · Работа и енергия · Момент на сила и лостове

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Quarterfinal 2005.pdf · официални решения: Quarterfinal 2005 Solutions.pdf